knitr::opts_chunk$set(echo = TRUE) library(tinytex) library(Deriv) library(reticulate) ## Warning: package 'reticulate' was built under R version 3.6.2 Sys.setenv(RETICULATE_PYTHON = "/Users/dlofland/anaconda3/bin/python") Source files: [https://github.com/djlofland/DATA605_S2020/tree/master/] ## Problem 1 Find the equation of the regression line for the given points. Round any final values to the nearest hundredth, if necessary. $( 5.6, 8.8 ), ( 6.3, 12.4 ), ( 7, 14.8 ), ( 7.7, 18.2 ), ( 8.4, 20.8 )$ x_values <- c(5.6, 6.3, 7.0, 7.7, 8.4) y_values <- c(8.8, 12.4, 14.8, 18.2, 20.8) x_bar <- mean(x_values) y_bar <- mean(y_values) sd_x <- sd(x_values) sd_y <- sd(y_values) r <- cor(y_values, x_values) m <- r * (sd_y/sd_x) b <- y_bar - m * x_bar print(paste('y = ', m, ' * x + ', b, sep='')) ## [1] "y = 4.25714285714286 * x + -14.8" plot(x_values, y_values) abline(b, m) ## Problem 2 ### Analytic Solution Find all local maxima, local minima, and saddle points for the function given below. Write your answer(s) in the form $$( x, y, z )$$. Separate multiple points with a comma. $f(x, y) = 24x - 6xy^2 - 8y^3$ Partial derivative with respect to $$x$$: $\frac{\partial{}}{\partial{x}}(24x - 6xy^2 - 8y^3) = 24 - 6y^2$ Partial derivative with respect to $$y$$: $\frac{\partial{}}{\partial{y}}(24x - 6xy^2 - 8y^3) = -12xy - 24y^2$ Set both partial derivatives to 0 and solve simulatneously: $24 - 6y^2 = 0\text{ and }-12xy - 24y^2 = 0\\ 6y^2 = 24\\ y^2 = 4, y = \pm2\\ -12xy - 24y^2 = 0\\ -12x(2) - 24(2^2) = 0\\ -24x - 96 = 0, x=-4\\ -12x(-2) - 24(-2)^2 = 0\\ 24x - 96 = 0, x = 4\\ (2, =4), (-2, 4)$ So (2, -4) and (-2, -4) are critical points. We need to second partial derivatives to determine whether we have minima, maxima ro saddle points. $f_{xx}(x, y) = \frac{\partial{}^2}{\partial{x}}(24x - 6xy^2 - 8y^3) = \frac{\partial{}}{\partial{x}}(24 - 6y^2) = 0\\ f_{yy}(x, y) = \frac{\partial{}^2}{\partial{y}}(24x - 6xy^2 - 8y^3) = \frac{\partial{}}{\partial{y}}(-12xy - 24y^2) = -12x-48y\\ f_{xy}(x, y) = \frac{\partial{}}{\partial{x}}(24x - 6xy^2 - 8y^3) = \frac{\partial{}}{\partial{y}}(24 - 6y^2) = -12y\\ f_{yx}(x, y) = \frac{\partial{}}{\partial{y}}(24x - 6xy^2 - 8y^3) = \frac{\partial{}}{\partial{x}}(-12xy - 24y^2) = -12y$ We examine the determinant $$D(x,y)$$ of the Hessian of $$f$$ at each critical point. $D(a, b) = f_{xx}(a,b)f_{yy}(a, b) - (f_{xy}(a,b))^2$ Points (2,-4), (-2, 4): $D(2,-4) = 0 - (-12*-4)^2 = -(48^2)\\ D(-2, 4) = 0 - (-12*4)(-12*4) = -(-48^2)$ • Since $$D(2,-4) < 0$$, point (2, -4) is a saddle point • Since $$D(-2, 4) < 0$$, point (-2, 4) is a saddle point • there are no maxima or minima ### Solution with R f <- function(x, y) {24*x - 6*x*y^2 - 8*y^3} (f_x <- Deriv(f, "x")) ## function (x, y) ## 24 - 6 * y^2 (f_y <- Deriv(f, "y")) ## function (x, y) ## -(y * (12 * x + 24 * y)) (f_xx <- Deriv(f_x, "x")) ## function (x, y) ## 0 (f_yy <- Deriv(f_y, "y")) ## function (x, y) ## -(12 * x + 48 * y) (f_xy <- Deriv(f_x, "y")) ## function (x, y) ## -(12 * y) ### Solution with Python import sympy as sp from sympy import symbols, exp x, y = symbols('x y') f = 24*x - 6*x*y**2 - 8*y**3 f_x = f.diff(x) f_y = f.diff(y) f_xx = f.diff(x, 2) f_yy = f.diff(y, 2) f_xy = f.diff(x, y) print('f: ', f) ## f: -6*x*y**2 + 24*x - 8*y**3 print('f_x: ', f_x) ## f_x: -6*y**2 + 24 print('f_y: ', f_y) ## f_y: -12*x*y - 24*y**2 print('f_xx: ', f_xx) ## f_xx: 0 print('f_yy: ', f_yy) ## f_yy: -12*(x + 4*y) print('f_xy: ', f_xy) # Find Critical Points ## f_xy: -12*y print('Critical Points: ', sp.solvers.solve([f_x, f_y], x, y)) # Use Determinant and Hessian to determine if minima, max or saddle ## Critical Points: [(-4, 2), (4, -2)] D = f_xx*f_yy - (f_xy)**2 print('D(-4, 2) = ', D.subs([(x, -4), (y, 2)])) ## D(-4, 2) = -576 print('D(2, -4) = ', D.subs([(x, 4), (y, -2)])) ## D(2, -4) = -576 Since D(-4, 2) and D(4, -2) are both < 0, both points are saddle points. There are no minima or maxima. ## Problem 3 A grocery store sells two brands of a product, the “house” brand and a “name” brand. The manager estimates that if she sells the “house” brand for $$x$$ dollars and the “name” brand for $$y$$ dollars, she will be able to sell $$81 - 21x + 17y$$ units of the “house” brand and $$40 + 11x - 23y$$ units of the “name” brand. ### Step 1. Find the revenue function R ( x, y ). $x=Price_{house}\\ y=Price_{name}\\ Units_{house} = 40 + 11x - 23y\\ Units_{name} = 81 - 21x + 17y\\ Revenue_{house} = Price_{house} * Units_{house} = x(40 + 11x - 23y)\\ Revenue_{name} = Price_{name} * Units_{name} = y(81 - 21x + 17y)\\ Revenue_{total} = Revenue_{house} + Revenue_{name}\\ Revenue_{total} = 40x + 44x^2 - 23xy + 81y - 21xy + 17y^2\\ R(x, y) = Revenue_{total} = 44x^2 + 40x - 44xy + 81y + 17y^2$ ### Step 2. What is the revenue if she sells the “house” brand for$2.30 and the “name” brand for \$4.10?

R_xy <- function(x, y) {44*x^2 + 40*x - 44*x*y + 81*y + 17*y^2}

R_xy(2.30, 4.10)
## [1] 527.71

## Problem 4

A company has a plant in Los Angeles and a plant in Denver. The firm is committed to produce a total of 96 units of a product each week. The total weekly cost is given by

$C(x, y) = \frac{1}{6}x^2 + \frac{1}{6}y^2 + 7x + 25y + 700$

where $$x$$ is the number of units produced in Los Angeles and $$y$$ is the number of units produced in Denver. How many units should be produced in each plant to minimize the total weekly cost?

### Symbolic Solution (Python)

from sympy import solve_linear

# ----- Symbolic Math Solution -----
x, y = symbols('x y')

f = 1/6*x**2 + 1/6*y**2 + 7*x + 25*y + 700
print('f: ', f)

# since x+y=96, we solve for x=96-y and substitute 96-y for x in f
# then solve for dC with respect to the single variable y
## f:  0.166666666666667*x**2 + 7*x + 0.166666666666667*y**2 + 25*y + 700
f_y = f.subs(x, 96-y).diff(y)
print('f_y: ', f_y)
## f_y:  0.666666666666667*y - 14.0
final_y = round(solve_linear(f_y)[1])
final_x = round(96 - final_y)

print('Symbolic Solution  ... x: ', final_x, ' y: ', final_y)
## Symbolic Solution  ... x:  75  y:  21

### Numerical Solution (Python)

Brute Force Method

# ----- Numerical Solution -----

min_cost = 1e6
x_soln = 0
y_soln = 0

for i in range(0,97):
c = f.subs([(x, i), (y, 96-i)])
if c < min_cost:
min_cost = c
x_soln = i
y_soln = 96-x_soln

print('Numerical Solution ... x: ', x_soln, ' y: ', y_soln, ' cost: ', round(c, 2))
## Numerical Solution ... x:  75  y:  21  cost:  2908.0

## Problem 5

Evaluate the double integral on the given region.

$\iint_{R}e^{(8x+3y)}dA;\ \ \ R: 2 \le x \le 4\text{ and }2\le y \le 4$

### Analytical Solution

$\int_{2}^{4}\int_{2}^{4}e^{(8x+3y)}\ dx\ dy\\ u = (8x + 3y);\ \ \ du = 8\ dx\\ \int_{2}^{4} \frac{1}{8}\int_{2}^{4}e^u\ du\ dy\\ \int_{2}^{4} \frac{e^{8x+3y}}{8} |_{2}^{4}\ dy\\ \int_{2}^{4} \frac{e^{3y+32}}{8} - \frac{e^{3y+16}}{8}\ dy\\ \int_{2}^{4} \frac{e^{3y+32}}{8}\ dy - \int_{2}^{4} \frac{e^{3y+16}}{8}\ dy\\ u=(3y+32);\ \ \ du=3\ dy\\ \frac{1}{8}\frac{1}{3}\int_{2}^{4} e^{u}\ du - \int_{2}^{4} \frac{e^{3y+16}}{8}\ dy\\ \frac{e^{3y+32}}{24}|_{2}^{4} - \int_{2}^{4} \frac{e^{3y+16}}{8}\ dy\\ \frac{e^{44}-e^{38}}{24} - \int_{2}^{4} \frac{e^{3y+16}}{8}\ dy\\ u=(3y+16);\ \ \ du=3\ dy\\ \frac{e^{44}-e^{38}}{24} - \frac{1}{8}\frac{1}{3}\int_{2}^{4} e^{u}\ du\\ \frac{e^{44}-e^{38}}{24} -\frac{e^{3y+16}}{24}|_{2}^{4}\\ \frac{e^{44}-e^{38}}{24} - \frac{e^{28} - e^{22}}{24}\\ \frac{e^{44}-e^{38} - e^{28} + e^{22}}{24}$

### Symbolic Solution (Python)

from sympy import integrate, Integral
x, y = symbols('x y')

f = exp(8*x + 3*y)
f
## exp(8*x + 3*y)
integrate(f, (x, 2, 4), (y, 2, 4))
## -(-1 + exp(16))*exp(22)/24 + (-1 + exp(16))*exp(28)/24

Note: By rearrangement and factoring out an $$e^{22}$$ and $$e^{28}$$, we get the same answer as with the analytical appraoch above.